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Math Help - [SOLVED] Sine Question

  1. #1
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    [SOLVED] Sine Question

    On a sunny day, a tower casts a shadow 35.2m long. At the same time, a 1.3m parking meter that is nearby casts a shadow 1.8m long. How high is the tower to the nearest tenth of a metre?

    Triangle formed by parking meter:
    c^2 = a^2 + b^2
    c^2 = 1.3^2 + 1.8^2
    c^2 = \frac{493}{100}
    c = \frac{\sqrt{493}}{10}

    SinA = \frac{O}{H}
    SinA = \frac{1.3}{\frac{\sqrt{493}}{10}}
    Sin^{-1}\frac{1.3}{\frac{\sqrt{493}}{10}} = A
    35.84^{\circ} = A

    Triangle formed by tower (Total triangle):
    \frac{a}{SinA} = \frac{b}{SinB}
    \frac{a}{Sin35.84} = \frac{37}{Sin54.16}
    a = Sin35.84(\frac{37}{Sin54.16})
    a = 26.7m

    *The 37m value was obtained by adding 35.2m and 1.8m.
    *The 54.16^{\circ} value was obtained by subtracting 90deg and 35.84deg from 180deg.

    Therefore, the height of the tower is 26.7m.

    Textbook answer: 25.4m.

    Where did I make the mistake?
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  2. #2
    A riddle wrapped in an enigma
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    Hi RogueDemon,

    Actually, you made this way harder than it is.

    You have two similar right triangles.

    Set up a proportion using the ratios of the heights and shadows.

    \frac{bldg \:\: ht.}{1.3}=\frac{35.2}{1.8}
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  3. #3
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    Thanks very much!
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